= Solution
Let $w$ denote the value of the <complex logarithm> used in the equation. Factoring $w^2+\pi^2/4$ gives $w=i\pi/2$ or $w=-i\pi/2$. Since $z=e^w$, \b[the complete answer is]
$$
\boxed{z=i\quad\text{or}\quad z=-i.}
$$
The <principal complex logarithm> takes exactly these values at $i$ and $-i$, so both are solutions with that convention. If the <complex logarithm> is interpreted as multivalued, the same answer holds in the sense that a logarithm value satisfies the equation. The equation is not an assertion that every logarithm value of these numbers has the stated square.
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